Question

I need help with my ap cal hw..pictures are available..both questions pertain to the same info. thanks! (its 1in the morning. this is due today 1st period.HELP!!)please. im confused. really am.

Answer 1

For the first one, you have horizontal tangents when y=constant.
The same applies for vertical tangents except at x=constant. That's why they are tangent lines, they touch the graph at one point and one point only. As such, you can't have a vertical at y=; you can only have a vertical line at x=. The same thing applies for horizontal tangent lines. I think you need to reread the question's options.

Answer 2

By the way, do you know how to solve for f(x)?

Answer 3

no. i do not know how to do any of the questions i posted.

Answer 4

divide by y and multiply by dx. Then integrate each side separately.
the integral of dy/y is ln(abs(y)). Then integrate the cosine. Then use the e to cancel out the ln on the left to get y=

Answer 5

...lol

Answer 6

I take it you're a sophomore, maybe junior? Differential equations is AP calc AB.

Answer 7

Look in your textbook.

Answer 8

\[{dy\over y} = {1\over 2}\cos x dx\]
\[{\int\limits_{?}^{?}\frac {dy} y} = \int\limits_{?}^{?} {1\over 2}\cos xdx\]
\[\ln \left| y \right| = {\frac 12}\sin x+C\]

Answer 9

Do you follow?

Answer 10

Sorry, i gtg. I'll finish off with the final step.

Answer 11

\[y = Ce ^{{\frac 1 2}\sin x}\]

Now just plug in the f(0)=2

Answer 12

Message me if you're still confused. Sorry.

Answer 13

thanks soo much

Answer 14

after i plug in im done with that part right?

Answer 15

At commented solution is attached.